1 / 39

CS252 Graduate Computer Architecture Lecture 15 Multiprocessor Networks March 14 th , 2011

CS252 Graduate Computer Architecture Lecture 15 Multiprocessor Networks March 14 th , 2011. John Kubiatowicz Electrical Engineering and Computer Sciences University of California, Berkeley http://www.eecs.berkeley.edu/~kubitron/cs252. Topological Properties.

leighj
Télécharger la présentation

CS252 Graduate Computer Architecture Lecture 15 Multiprocessor Networks March 14 th , 2011

An Image/Link below is provided (as is) to download presentation Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author. Content is provided to you AS IS for your information and personal use only. Download presentation by click this link. While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server. During download, if you can't get a presentation, the file might be deleted by the publisher.

E N D

Presentation Transcript

  1. CS252Graduate Computer ArchitectureLecture 15Multiprocessor NetworksMarch 14th, 2011 John Kubiatowicz Electrical Engineering and Computer Sciences University of California, Berkeley http://www.eecs.berkeley.edu/~kubitron/cs252

  2. Topological Properties • Routing Distance - number of links on route • Diameter - maximum routing distance • Average Distance • A network is partitioned by a set of links if their removal disconnects the graph cs252-S11, Lecture 15

  3. Interconnection Topologies • Class of networks scaling with N • Logical Properties: • distance, degree • Physical properties • length, width • Fully connected network • diameter = 1 • degree = N • cost? • bus => O(N), but BW is O(1) - actually worse • crossbar => O(N2) for BW O(N) • VLSI technology determines switch degree cs252-S11, Lecture 15

  4. Example: Linear Arrays and Rings • Linear Array • Diameter? • Average Distance? • Bisection bandwidth? • Route A -> B given by relative address R = B-A • Torus? • Examples: FDDI, SCI, FiberChannel Arbitrated Loop, KSR1 cs252-S11, Lecture 15

  5. Example: Multidimensional Meshes and Tori • n-dimensional array • N = kn-1 X ...X kO nodes • described by n-vector of coordinates (in-1, ..., iO) • n-dimensional k-ary mesh: N = kn • k = nÖN • described by n-vector of radix k coordinate • n-dimensional k-ary torus (or k-ary n-cube)? 3D Cube 2D Grid 2D Torus cs252-S11, Lecture 15

  6. On Chip: Embeddings in two dimensions • Embed multiple logical dimension in one physical dimension using long wires • When embedding higher-dimension in lower one, either some wires longer than others, or all wires long 6 x 3 x 2 cs252-S11, Lecture 15

  7. Trees • Diameter and ave distance logarithmic • k-ary tree, height n = logk N • address specified n-vector of radix k coordinates describing path down from root • Fixed degree • Route up to common ancestor and down • R = B xor A • let i be position of most significant 1 in R, route up i+1 levels • down in direction given by low i+1 bits of B • H-tree space is O(N) with O(ÖN) long wires • Bisection BW? cs252-S11, Lecture 15

  8. Fat-Trees • Fatter links (really more of them) as you go up, so bisection BW scales with N cs252-S11, Lecture 15

  9. Butterflies • Tree with lots of roots! • N log N (actually N/2 x logN) • Exactly one route from any source to any dest • R = A xor B, at level i use ‘straight’ edge if ri=0, otherwise cross edge • Bisection N/2 vs N (n-1)/n(for n-cube) building block 16 node butterfly cs252-S11, Lecture 15

  10. k-ary n-cubes vs k-ary n-flies • degree n vs degree k • N switches vs N log N switches • diminishing BW per node vs constant • requires locality vs little benefit to locality • Can you route all permutations? cs252-S11, Lecture 15

  11. Benes network and Fat Tree • Back-to-back butterfly can route all permutations • What if you just pick a random mid point? cs252-S11, Lecture 15

  12. Hypercubes • Also called binary n-cubes. # of nodes = N = 2n. • O(logN) Hops • Good bisection BW • Complexity • Out degree is n = logN correct dimensions in order • with random comm. 2 ports per processor 0-D 1-D 2-D 3-D 4-D 5-D ! cs252-S11, Lecture 15

  13. Some Properties • Routing • relative distance: R = (b n-1 - a n-1, ... , b0 - a0 ) • traverse ri = b i - a i hopsin each dimension • dimension-order routing? Adaptive routing? • Average Distance Wire Length? • n x 2k/3 for mesh • nk/2 for cube • Degree? • Bisection bandwidth? Partitioning? • k n-1 bidirectional links • Physical layout? • 2D in O(N) space Short wires • higher dimension? cs252-S11, Lecture 15

  14. The Routing problem: Local decisions • Routing at each hop: Pick next output port! cs252-S11, Lecture 15

  15. How do you build a crossbar? cs252-S11, Lecture 15

  16. Input buffered switch • Independent routing logic per input • FSM • Scheduler logic arbitrates each output • priority, FIFO, random • Head-of-line blocking problem • Message at head of queue blocks messages behind it cs252-S11, Lecture 15

  17. Output Buffered Switch • How would you build a shared pool? cs252-S11, Lecture 15

  18. Properties of Routing Algorithms • Routing algorithm: • R: N x N -> C, which at each switch maps the destination node nd to the next channel on the route • which of the possible paths are used as routes? • how is the next hop determined? • arithmetic • source-based port select • table driven • general computation • Deterministic • route determined by (source, dest), not intermediate state (i.e. traffic) • Adaptive • route influenced by traffic along the way • Minimal • only selects shortest paths • Deadlock free • no traffic pattern can lead to a situation where packets are deadlocked and never move forward cs252-S11, Lecture 15

  19. Example: Simple Routing Mechanism • need to select output port for each input packet • in a few cycles • Simple arithmetic in regular topologies • ex: Dx, Dy routing in a grid • west (-x) Dx < 0 • east (+x) Dx > 0 • south (-y) Dx = 0, Dy < 0 • north (+y) Dx = 0, Dy > 0 • processor Dx = 0, Dy = 0 • Reduce relative address of each dimension in order • Dimension-order routing in k-ary d-cubes • e-cube routing in n-cube cs252-S11, Lecture 15

  20. Communication Performance • Typical Packet includes data + encapsulation bytes • Unfragmented packet size S = Sdata+Sencapsulation • Routing Time: • Time(S)s-d = overhead + routing delay + channel occupancy + contention delay • Channel occupancy = S/b = (Sdata+ Sencapsulation)/b • Routing delay in cycles (D): • Time to get head of packet to next hop • Contention? cs252-S11, Lecture 15

  21. Store&Forward vs Cut-Through Routing Time: h(S/b + D/) vs S/b + h D/ OR(cycles): h(S/w + D) vs S/w + h D • what if message is fragmented? • wormhole vs virtual cut-through cs252-S11, Lecture 15

  22. Contention • Two packets trying to use the same link at same time • limited buffering • drop? • Most parallel mach. networks block in place • link-level flow control • tree saturation • Closed system - offered load depends on delivered • Source Squelching cs252-S11, Lecture 15

  23. Bandwidth • What affects local bandwidth? • packet density: b x Sdata/S • routing delay: b x Sdata /(S + wD) • contention • endpoints • within the network • Aggregate bandwidth • bisection bandwidth • sum of bandwidth of smallest set of links that partition the network • total bandwidth of all the channels: Cb • suppose N hosts issue packet every M cycles with ave dist • each msg occupies h channels for l = S/w cycles each • C/N channels available per node • link utilization for store-and-forward:r = (hl/M channel cycles/node)/(C/N) =Nhl/MC< 1! • link utilization for wormhole routing? cs252-S11, Lecture 15

  24. Saturation cs252-S11, Lecture 15

  25. How Many Dimensions? • n = 2 or n = 3 • Short wires, easy to build • Many hops, low bisection bandwidth • Requires traffic locality • n >= 4 • Harder to build, more wires, longer average length • Fewer hops, better bisection bandwidth • Can handle non-local traffic • k-ary n-cubes provide a consistent framework for comparison • N = kn • scale dimension (n) or nodes per dimension (k) • assume cut-through cs252-S11, Lecture 15

  26. Traditional Scaling: Latency scaling with N • Assumes equal channel width • independent of node count or dimension • dominated by average distance cs252-S11, Lecture 15

  27. Average Distance • but, equal channel width is not equal cost! • Higher dimension => more channels ave dist = n(k-1)/2 cs252-S11, Lecture 15

  28. Dally Paper: In the 3D world • For N nodes, bisection area is O(N2/3 ) • For large N, bisection bandwidth is limited to O(N2/3 ) • Bill Dally, IEEE TPDS, [Dal90a] • For fixed bisection bandwidth, low-dimensional k-ary n-cubes are better (otherwise higher is better) • i.e., a few short fat wires are better than many long thin wires • What about many long fat wires? cs252-S11, Lecture 15

  29. Logarithmic Delay Linear Delay Dally paper (con’t) • Equal Bisection,W=1 for hypercube  W= ½k • Three wire models: • Constant delay, independent of length • Logarithmic delay with length (exponential driver tree) • Linear delay (speed of light/optimal repeaters) cs252-S11, Lecture 15

  30. Equal cost in k-ary n-cubes • Equal number of nodes? • Equal number of pins/wires? • Equal bisection bandwidth? • Equal area? • Equal wire length? What do we know? • switch degree: n diameter = n(k-1) • total links = Nn • pins per node = 2wn • bisection = kn-1 = N/k links in each directions • 2Nw/k wires cross the middle cs252-S11, Lecture 15

  31. Latency for Equal Width Channels • total links(N) = Nn cs252-S11, Lecture 15

  32. Latency with Equal Pin Count • Baseline n=2, has w = 32 (128 wires per node) • fix 2nw pins => w(n) = 64/n • distance up with n, but channel time down cs252-S11, Lecture 15

  33. Latency with Equal Bisection Width • N-node hypercube has N bisection links • 2d torus has 2N 1/2 • Fixed bisection w(n) = N 1/n / 2 = k/2 • 1 M nodes, n=2 has w=512! cs252-S11, Lecture 15

  34. Larger Routing Delay (w/ equal pin) • Dally’s conclusions strongly influenced by assumption of small routing delay • Here, Routing delay =20 cs252-S11, Lecture 15

  35. Saturation • Fatter links shorten queuing delays cs252-S11, Lecture 15

  36. Discuss of paper: Virtual Channel Flow Control • Basic Idea: Use of virtual channels to reduce contention • Provided a model of k-ary, n-flies • Also provided simulation • Tradeoff: Better to split buffers into virtual channels • Example (constant total storage for 2-ary 8-fly): cs252-S11, Lecture 15

  37. When are virtual channels allocated? • Two separate processes: • Virtual channel allocation • Switch/connection allocation • Virtual Channel Allocation • Choose route and free output virtual channel • Really means: Source of link tracks channels at destination • Switch Allocation • For incoming virtual channel, negotiate switch on outgoing pin Hardware efficient design For crossbar cs252-S11, Lecture 15

  38. Reducing routing delay: Express Cubes • Problem: Low-dimensional networks have high k • Consequence: may have to travel many hops in single dimension • Routing latency can dominate long-distance traffic patterns • Solution: Provide one or more “express” links • Like express trains, express elevators, etc • Delay linear with distance, lower constant • Closer to “speed of light” in medium • Lower power, since no router cost • “Express Cubes: Improving performance of k-ary n-cube interconnection networks,” Bill Dally 1991 • Another Idea: route with pass transistors through links cs252-S11, Lecture 15

  39. Summary • Network Topologies: • Fair metrics of comparison • Equal cost: area, bisection bandwidth, etc • Routing Algorithms restrict set of routes within the topology • simple mechanism selects turn at each hop • arithmetic, selection, lookup • Virtual Channels • Adds complexity to router • Can be used for performance • Can be used for deadlock avoidance Topology Degree Diameter Ave Dist Bisection D (D ave) @ P=1024 1D Array 2 N-1 N / 3 1 huge 1D Ring 2 N/2 N/4 2 2D Mesh 4 2 (N1/2 - 1) 2/3 N1/2 N1/2 63 (21) 2D Torus 4 N1/2 1/2 N1/2 2N1/2 32 (16) k-ary n-cube 2n nk/2 nk/4 nk/4 15 (7.5) @n=3 Hypercube n =log N n n/2 N/2 10 (5) cs252-S11, Lecture 15

More Related